Proving that $a \mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb Z$ [duplicate]
Given $m \in \mathbb{Z}$, let $m\mathbb{Z}$ denote the set of integer multiples of $m$, i.e. $m\mathbb{Z} := \{mk\mid k \in \mathbb{Z}\}$. Now let $a,b \in \mathbb{Z}$ with $a,b$ not both $0$. Prove that $a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb{Z}$.
I am trying to write a proof for this, but I am unsure of what method to use. Also I am confused by $mk\mid k$, because wouldn't $m=1$ for this to be true.
Hint:
Step 1: Can you prove that $\operatorname{lcm}(a,b)\in a\mathbb{Z}\cap b\mathbb{Z}$?
Step 2: Can you prove that if $a$ and $b$ both divide $c$, then $c\in a\mathbb{Z}\cap b\mathbb{Z}$?
Step 3: How do steps 1 and 2, together, imply your result?
First of all, to clear up your confusion:
$$m\mathbb Z=\{mk\mid k\in\mathbb Z\}$$
does not mean that $mk$ divides $k$. The vertical line can be read as "so that" or "where". It means that the set $m\mathbb Z$ is the set of numbers in the form of $mk$, where $k$ is any element of $\mathbb Z$.
That said, to prove that $a\mathbb Z\cap b\mathbb Z=\operatorname{lcm}(a,b)\mathbb Z$, you need to prove that
- if $x\in a\mathbb Z\cap b\mathbb Z$, then $x\in \operatorname{lcm}(a,b)\mathbb Z$.
- if $x\in\operatorname{lcm}(a,b)\mathbb Z$, then $x\in a\mathbb Z\cap b\mathbb Z$.
I give another demonstration you might find useful, which allows you to reach the result without double implication, using only the definition of $lcm(a,b)$.
Definition: $d \in \Bbb Z$ is defined least common multiple of $a$ and $b$, in symbols $d := lcm(a,b)$ if:
- $\exists m,n \in \Bbb Z$ such that $am = d = bn$, otherwise written as $a|d$ and $b|d$
- If $c \in \Bbb Z$ is such that $a|c$ and $b|c$, then $d|c$
So we can write $$a\Bbb Z \cap b\Bbb Z =_{(1)} \left \{ x \in \Bbb Z : a|x, b|x \right \} =_{(2)} \left \{ x \in \Bbb Z : lcm(a,b)|x \right \} =_{(3)} lcm(a,b)\Bbb Z$$
Where:
- $=_{(1)}, =_{(3)}$ follows from the definition of $a\Bbb Z \cap b\Bbb Z$ and $lcm(a,b)\Bbb Z$
- $=_{(2)}$ follows from the point $2$ of the definition of $lcm(a,b)$